How do you rewrite (sin x - cos x)(sin x + cos x)? Thank you.

You're probably used to dealing with this only in quadratics, but the expression is in the difference of squares pattern

#(a-b)(a+b)=a^2-b^2#

where #color(red)(a=sinx)# and #color(blue)(b=cosx)#

We can just plug in our values for #a# and #b# into our difference of squares expression, and we get:

#color(red)((sinx))^2-color(blue)((cosx))^2#

Which can be rewritten as

#sin^2x-cos^2x#

If you don't believe me, we can FOIL this expression to make sure:

With FOIL, we multiply the first, outside, inside and last terms and add the result. Thus, we have:

• First terms: #sinx*sinx=color(red)(sinx^2)#
• Outside terms: #sinx*cosx=sinxcosx#
• Inside terms: #sinx*-cosx=-sinxcosx#
• Last terms: #-cosx*cosx=-color(blue)(cosx^2)#

Now we have

#color(red)((sinx))^2+cancel(sinxcosx-sinxcosx)-color(blue)((cosx))^2#

The middle terms obviously cancel out, and we can rewrite this as

#sin^2x-cos^2x#

The key realization is that the original expression in question was in a difference of squares pattern.

If you have a #(a-b)(a+b)# pattern, you can always use the difference of squares identity to quickly simplify it.

Hope this helps!

#f(x) = (sin x - cos x)(sin x + cos x) = (sin^2 x - cos^2 x) #
Reminder of trig identity:
#cos 2x = cos^2 x - sin^2 x#
Finally,
#f(x) = - cos 2x#